Price And Productivity Determination Under Perfect Rivalry

Will the industry manufacture the commodity if price of the commodity is $80?

Solution

An industry manufactures a commodity if price of the commodity surpass its minimum average
variable cost.

AVC = TVC = 150V – 20V^2
+ 2V^3
V V

= 150 – 20V
+ 2V^2 …..Equation (1)

AVC is minimised at the output level at which

d
(AVC) = 0
d
V

taking the derivative of AVC, we procure:

d
AVC = -20
+ 4V
d
V

Thus, AVC will be minim um when

-20
+ 4V = 0

4V = 20

V = 20
/ 4 = 5

Now substituting the value of V in the Equation for AVC we procure,

Minimum
AVC = 150
- 20 * 5 + 2 * (5)^2

= 150
- 100 + 50 = 100

Therefore, price of $80 of the product is less than the minimum average variable
cost the firm will not produce the product.

Illustration 63

An industry functioning in a purely rival environment is faced with a market price of
$125 per unit of the commodity. The industry’s aggregate short run cost function
is as follows: TC = 3000
+ 200V – 10V^2 + 2V^3.

Should the industry manufacture at this price in the short run?

If the market price is $150 per unit, what will aggregate profits or losses be
if the industry manufactures 5 units of productivity? Should the industry manufacture
at this price?

If the market price is greater than $150, should the industry manufacture at this
price?

Solution

An industry will continue manufacturing in the short run if price of the commodity
of $125 surpasses the minimum average variable cost. So we have first to ascertain
the minimum average variable cost.

It is to be noted that in the given cost function $3000 is fixed cost for the reason
that it does not incorporate any productivity component V. Therefore,

TVC = 200V – 10V^2
+ 2V^3

AVC = TVC = 200V – 10V^2
+ 2V^3
V V

AVC = 200 – 10V
+ 2V^2 …..Equation
(1)

To ascertain the level of productivity at which average variable cost is minimum, we
consider the first derivative of AVC function and set it equal to zero.

dAVC = -
10 + 4V
dV

Fixing dAVC = 0,
we have
dV

-
10 + 4V = 0

4V = 10

V = 10
/ 4 = 2.5

Substituting the value of V in the AVC function Equation (1), we have

Minimum AVC = 200 – 10
(2.5) + 2 (2.5)^2

= 200 – 25
+ 12.5 = 187.5

Since the price of $125 is less than the minimum average variable cost of $150,
the industry will not manufacture in the short run for the reason that it will
not even gain back variable costs.

If the market price of the commodity is $150, it may persist manufacturing
in the short run for the reason that it will be covering the variable costs
entirely though it will not be gaining back any part of the fixed costs and
thus incurring losses.

It may be noted that at price $150, the industry shall manufacture 2.5 units. This
can be known by equating this price with marginal cost which is the profit optimising
situation under perfect rivalry. Therefore,

MC = dTVC = 200 – 20V
+ 6V^2
dV

MC = Price = 150

200 – 20V
+ 6V^2 = 150

V = 2.5

ATC = 3000
+ 200V – 10V^2 + 2V^3
V

= 3000 +
200 – 10V + 2V^2
V

ATC at productivity 2.5 = 6000 +
200 – 10 (2.5) + 2 (2.5)^2
2.5

= 2400
+ 200 – 25 + 12.5

= 2587.5

Losses at price $150:

π = TR – TC

TR = P.V – ATC.V

= (150
*2.5) – (2587.5 * 2.5)

= 375 – 6468.75

= -
6093.75

Therefore, the industry will be incurring losses equal to $3000 at price $150
per unit that is losses are almost double the total fixed cost.

Illustration 64

Presume that revenue and total cost of an industry are provided by the equations:

A = 120V and F = 20 + 10V^2, where V is the volume of output.

Using the TR, TC method, ascertain the profit optimisation level and total profits of
the industry.

Solution

π = TR – TC

= 120V – 20 – 10V^2

Profits will be optimum at the level of productivity at which the first derivative of
total profit function = 0. Therefore,

dπ = 120 – 20V
dV

setting dπ equal to zero, we have,
dV

120 – 20V = 0

20V = 120

V = 120
/ 20 = 6

For the second order condition to be satisfied, the second order derivative of profit
function must be negative. Taking the second derivative of the profit function, we
have

d^2 π = -
20
dV^2

Therefore, the second order condition is fulfilled.

Substituting, V = 6 in the profit function, we have

π = 120*6 – 20 – 10
(6)^2

= 720 – 20 – 360

π = 340

Therefore, the profit optimising productivity will be 6 and that the total profit
of the industry will be 340.

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